(i) Using Newton's second law for both particles:

For particle A: \(T - 0.3g = 0.3a\)

For particle B: \(0.7g - T = 0.7a\)

Adding these equations gives:

\(0.7g - 0.3g = (0.7 + 0.3)a\)

\(0.4g = a\)

\(a = 4 \text{ m s}^{-2}\)

(ii) Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = -4 \text{ m s}^{-2}\):

\(0 = 1.6^2 - 2 \times 4 \times s\)

\(s = \frac{1.6^2}{2 \times 4} = 0.32 \text{ m}\)

For the total height, \(s_1 + s_2 = 0.32 + 0.128 = 0.448 \text{ m}\)

(iii) Using \(v = u + at\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = 4 \text{ m s}^{-2}\):

\(1.6 = 4t_1\)

\(t_1 = \frac{1.6}{4} = 0.4 \text{ s}\)

For the descent, \(t_2 = \frac{1.6}{10} = 0.16 \text{ s}\)

Total time \(t_1 + t_2 = 0.4 + 0.16 = 0.56 \text{ s}\)